108年調查三等考試-工程數學詳解

108年法務部調查局調查人員考試

考試別：調查人員
等別：三等考試
類 科 ：電子科學組
科 目：工程數學

解：
$$令\begin{cases} P(x,y)={y\over x^2}\\ Q(x,y)=-{1\over x}  \end{cases} \Rightarrow {\partial Q\over \partial x}-{\partial P\over \partial y}={1\over x^2}-{1\over x^2}=0 \Rightarrow  \oint_C{\left({y\over x^2}dx-{1\over x}dy\right)} = \int\int_R{0\,dA}=\bbox[red, 2pt]{0}$$

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解：
$$A=\left[ \begin{array}{rrr}

-2 & 4 & 2 \\

 6 & 3 &-3 \\

 2 & 9 &-5
\end{array} \right]
\Rightarrow det(A)=30+108-24-12+120-54=168\\

\Rightarrow A^{-1}={1\over det(A)}\left[ \begin{array}{rrr}

\left| \begin{matrix} 3& -3\\9&-5 \end{matrix} \right| & -\left| \begin{matrix} 4& 2\\9 &-5 \end{matrix} \right| & \left| \begin{matrix} 4& 2\\3 &-3 \end{matrix} \right| \\

-\left| \begin{matrix} 6& -3\\2&-5 \end{matrix} \right| & \left| \begin{matrix} -2& 2\\2&-5 \end{matrix} \right| & -\left| \begin{matrix} -2& 2\\6 &-3 \end{matrix} \right|
  \\
 \left| \begin{matrix} 6& 3\\2& 9 \end{matrix} \right| & -\left| \begin{matrix} -2& 4\\2 & 9 \end{matrix} \right| & \left| \begin{matrix} -2& 4\\6 & 3 \end{matrix} \right|

\end{array} \right]\\

={1\over 168}\left[ \begin{array}{rrr}

12 & 38 & -18 \\
 24 & 6 & 6 \\
 48 & 26 &-30

\end{array} \right] = \bbox[ red, 2pt]{\bbox[red, 2pt]{\left[ \begin{array}{rrr}
1/14 & 19/84 & -3/28 \\
1/7 & 1/28 & 1/28 \\
2/7 & 13/84 &-5/28
\end{array} \right] }}$$

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解：
$$f(z)={\sin^2{z} \over z^2(z^2+4)} ={\sin^2{z} \over z^2(z+2i)(z-2i)} \Rightarrow \begin{cases}\text{Res}(f,0)= \left. {d\over dz} {\sin^2{z} \over z^2+4}\right|_{z=0}=\left. {2\sin{z}\cos{z}\over z^2+4}- {2z\sin^2{z} \over (z^2+4)^2}\right|_{z=0}=0\\\text{Res}(f,2i)= \left .{\sin^2{z} \over z^2(z+2i)} \right|_{z=2i}={ i\sin^2{2i} \over 16}\end{cases}\\
\Rightarrow \oint_{\Gamma}{f(z)\,dz}=2\pi i\times(\text{Res}(f,0)+\text{Res}(f,2i))= 2\pi i \times { i\sin^2{2i} \over 16}= \bbox[red, 2pt]{-{\pi \sin^2{(2i)}\over 8}}$$

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解：$$F(\omega)=\int_{-\infty}^\infty{f(t)e^{-i\omega t}\,dt} =\int_{-\infty}^\infty {e^{-2t^2}e^{-i\omega t}\,dt} \Rightarrow {d\over d\omega}F(\omega) = {-i}\int_{-\infty}^\infty {te^{-2t^2}e^{-i\omega t}\,dt}\\
={-i}\left( \left. \left[ -{1\over 4}e^{-2t^2-i\omega t} \right] \right|_{-\infty}^\infty -{i\omega \over 4}\int_{-\infty}^\infty {e^{-2t^2}e^{-i\omega t}\,dt} \right) = {-i}\left( 0 -{i\omega \over 4}\int_{-\infty}^\infty {e^{-2t^2}e^{-i\omega t}\,dt} \right)\\
=-{\omega\over 4}\int_{-\infty}^\infty {e^{-2t^2}e^{-i\omega t}\,dt} =-{\omega\over 4}F(\omega) \Rightarrow {d\over d\omega}F(\omega)=-{\omega\over 4}F(\omega) \\ \Rightarrow {dF(\omega) \over F(\omega)} ={-w\over 4}d\omega \Rightarrow \ln{|F(\omega)|}=-{1\over 8}\omega^2+C_1 \Rightarrow F(\omega)=C_2e^{-\omega^2/8}\\
F(0)= \int_{-\infty}^\infty {e^{-2t^2}e^{0}\,dt} =\int_{-\infty}^\infty {e^{-2t^2}\,dt} \Rightarrow F(0)\times F(0)= \left(\int_{-\infty}^\infty {e^{-2x^2}dx}\right) \left(\int_{-\infty}^\infty{e^{-2y^2}dy}\right) \\= \int_{-\infty}^\infty\int_{-\infty}^\infty {e^{-2(x^2+y^2)}dydx} = \int_0^{2\pi}\int_0^\infty {re^{-2r^2}drd\theta}, 其中
\begin{cases} x=r\cos{\theta} \\y=r\sin{\theta}\end{cases}\\ \Rightarrow F(0)\times F(0) = \int_0^{2\pi}\left. \left[-{1\over 4}e^{-2r^2} \right]\right|_0^\infty\,d\theta =\int_0^{2\pi} {{1\over 4}d\theta} = {\pi \over 2} \Rightarrow F(0) =\sqrt{\pi \over 2} =C_2\\ \Rightarrow \bbox[red, 2pt]{F(\omega)= \sqrt{\pi \over 2}e^{-\omega^2/8}}$$

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解：$$f_X(x)= {1\over 2\pi},\forall x\in [-\pi, \pi] \Rightarrow F_X(x)=\int_{-\pi}^x{1\over 2\pi}dx= {x+\pi \over 2\pi} \Rightarrow F_X(x)= \begin{cases} 1 & x\ge \pi \\{x+\pi \over 2\pi} & -\pi\le x\le \pi\\ 0 &x \le -\pi  \end{cases}\\
F_Y(y)= P(Y\le y) =P(\cos{X} \le y) =P(X\le -\cos^{-1}{y})+P(X\ge \cos^{-1}{y}) \\ = P(X\le -\cos^{-1}{y})+1-P(X\le \cos^{-1}{y})= F_X(-\cos^{-1}{y})+ 1-F_X(\cos^{-1}{y})\\ =\begin{cases} 1 & y\ge 1 \\{-\cos^{-1}{y}+\pi \over 2\pi}+1-{\cos^{-1}{y}+\pi\over 2\pi} & -1\le y\le 1\\ 0 &y \le -1  \end{cases} =\begin{cases} 1 & y\ge 1 \\1-{\cos^{-1}{y}\over \pi} & -1\le y\le 1\\ 0 &y \le -1  \end{cases} \\\Rightarrow f_Y(y)={d\over dy}F_Y(y)= \bbox[red, 2pt]{\begin{cases} {1 \over \pi \sqrt{1-y^2}} & y\in [-1,1]\\ 0& 其他\end{cases}}$$

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